如何使用Serde进行反序列化时转换字段？

Question

我正在使用Serde反序列化一个XML文件，该文件的十六进制值为0x400作为字符串，我需要将其转换为值1024，作为u32。

我是否需要实现Visitor特征，以便将0x分开，然后将400从16位解码到10位？如果是这样，我该怎么做，以便基10整数的反序列化保持不变？

Answer 1

最简单的解决方案是使用Serde field attributedeserialize_with为您的字段设置自定义序列化函数。然后，您可以得到原始字符串和convert it as appropriate：

extern crate serde; 
#[macro_use] 
extern crate serde_derive; 
extern crate serde_json; 

use serde::{Deserialize, Deserializer}; 
use serde::de::Error; 

#[derive(Debug, Deserialize)] 
struct EtheriumTransaction { 
    #[serde(deserialize_with = "from_hex")] 
    account: u64, // hex 
    amount: u64, // decimal 
} 

fn from_hex<'de, D>(deserializer: D) -> Result<u64, D::Error> 
where 
    D: Deserializer<'de>, 
{ 
    let s: &str = Deserialize::deserialize(deserializer)?; 
    // do better hex decoding than this 
    u64::from_str_radix(&s[2..], 16).map_err(D::Error::custom) 
} 

fn main() { 
    let raw = r#"{"account": "0xDEADBEEF", "amount": 100}"#; 
    let transaction: EtheriumTransaction = 
     serde_json::from_str(raw).expect("Couldn't derserialize"); 
    assert_eq!(transaction.amount, 100); 
    assert_eq!(transaction.account, 0xDEAD_BEEF); 
} 

playground

从这里，这是一个很小的步骤，它推广到自己的类型，以便重新使用：

#[derive(Debug, Deserialize)] 
struct EtheriumTransaction { 
    account: Account, // hex 
    amount: u64,  // decimal 
} 

#[derive(Debug, PartialEq)] 
struct Account(u64); 

impl<'de> Deserialize<'de> for Account { 
    fn deserialize<D>(deserializer: D) -> Result<Self, D::Error> 
    where 
     D: Deserializer<'de>, 
    { 
     let s: &str = Deserialize::deserialize(deserializer)?; 
     // do better hex decoding than this 
     u64::from_str_radix(&s[2..], 16) 
      .map(Account) 
      .map_err(D::Error::custom) 
    } 
} 

playground

另请参阅：

• How to transform fields before serialization using serde?